How do you verify the identity $(tanx-secx)/(tanx+secx)=(tan^2x-1)/(sec^2x)$ ?

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How do you verify the identity $(tanx-secx)/(tanx+secx)=(tan^2x-1)/(sec^2x)$ ?

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Answer 1 · 2018-04-05T03:50:29

We have:

$(sinx/cosx - 1/cosx)/(sinx/cosx + 1/cosx) = (sin^2x/cos^2x- 1)/(1/cos^2x)$

$((sinx -1)/cosx)/((sinx +1)/cosx) = ((sin^2x - cos^2x)/cos^2x)/(1/cos^2x)$

$(sinx - 1)(sinx+ 1) = sin^2x - cos^2x$

$sin^2x - 2sinx - 1= sin^2x - cos^2x$

$sin^2x - 2sinx - (sin^2x + cos^2x) = sin^2x - cos^2x$

$-2sinx - cos^2x = sin^2x - cos^2x$

Since $-2sinx != sin^2x$ , this identity is false.

Hopefully this helps!

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How do you verify the identity $(tanx-secx)/(tanx+secx)=(tan^2x-1)/(sec^2x)$ ?

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